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STRUCTURE OF Cr-52 AND Cr-48
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( June 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ”(2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. STRUCTURE OF STABLE Cr-52 WITH S =0 ''' Comparing the Cr-52 with the Ca-40 we see that the structure of Cr-52 has the core of Ca-40 in which we add four deuterons and four extra neutrons with total spin S=0. In the following diagram you see the two deuterons of the fifth horizontal plane which belong to the nucleons of positive spins. Such deuterons like the p21n21 of S = +1, and p22n22 of S=+1 , give a total spin S=+2. Since the total spin of Cr-52 is S =0 in the diagram you see also the deuterons of the second horizontal plane with negative spins like the p23n23 of S=-1 and the p24n34 of S =-1, giving a total spin S= -2. That is the four deuterons contribute not only to the high symmetry but also to the spin S=0. However this structure gives the structure of the unstable Cr-48 because at the positions p21, p22, p23, and p24 we observe only three pn bonds per proton unable to overcome the pp repulsions of long range. Thus in order to have a stable structure , one must add four extra neutrons. In the diagram you see also the extra four neutrons, which fill blank positions between two protons like the n25(+1/2), the n26(+1/2), the n27(-1/2), and the n28(-1/2) of total spin S=0 . Of course they contribute to the increase of the pn bonds per proton able to overcome the pp repulsions of long range . For example the n26 makes the two np bond like the ( n26-p1) and the (n26-p24). So at the p24 we observe now the four pn bonds per proton as ( p24-n3), ( p24-n13), (p24-n24)and (p24-n26) able to overcome the pp repulsions of long range. '''HOW Cr-48 OF S = 0 DECAYS TO V-48 OF S = +4 Although Cr-48 has a structure of high symmetry it is unstable because it has the same number of protons and neutrons (24 protons and 24 neutrons). Note that the heavy nuclei with a large number of nucleons are unstable because the great number of protons contributes to the increase of the pp repulsions of long range which overcome the binding energy of pn bonds. Under the instability of Cr-48 the p24(-1/2) decays to n25(+1/2) of V-48. Note that such a transformation gives S =+1. Then in the absence of p24 the n24(-1/2) moves to the third plane as n26(+1/2) . Also this changing of spins gives S=+1. For understanding carefully the rearrangements of spins In my STRUCTURE OF Ti-48 and V-48 one sees that the spin S= +4 is due to two extra electrons of total spin S=+1 and of three deuterons like the p21n21 of S=+1 the p22n22 of S=+1 and the p23n23 of S= +1 of which p23 and n23 are not shown, because they are in front of n9 and p9 respectively . Thus during the transformation from Cr-48 to the V-48 the p23(-1/2) and n23(-1/2) of Cr-48 become p23(+1/2) and n23(+1/2). Note that this transformation increases the spin from S = 0 to S = +2 . In other words the spin S = 0 of the Cr-48 becomes S =+4 of the V-48 because the negative spins of p23, n23 p24 and n24, of the unstable Cr-48 turn to positive spins of p23, n23, n25 and n26 of V-48. STABLE STRUCTURE OF Cr-52 WITH S = 0 (The nucleons p19, n19, n20, and p20 are not shown here because they are behind the n6, p6, p8, and n8 respectively. Also the n17, p17, p18, and n18 are not shown here because they are in front of p5, n5, n7, and p7 respecively) ' ' n28……….p12..........n12 ' n11..........p11 …….n27 Sixth horizontal plane of 6 nucleons' ' p22.........n10..........p10..........n21' ' n22..........p9............n9..........p21 Fifth horizontal plane of 8 nucleons' '' '' n14..........p8............n8............p16 '' p14..........n7............p7...........n16 '' '' Fourth horizontal plane of 8 nucleons '' ' p13..........n6...........p6............n15' ' n13..........p5...........n5............p15 Third horizontal plane of 8 nucleons' ' ' '' '' n24……….p4............n4……….p23 ' p24………n3..........p3 ……… n23 Second horizontal plane of 8 nucleons' ' n2............p2……....n25' ' n26……...p1...........n1 First horizontal plane of 6 nucleons ''' ''' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' Category:Fundamental physics concepts